Chem. 1B Midterm 2 Version B Feb. 24, 2016
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1 Chem. B Midterm 2 Version B Feb. 24, 206 First initial of last name Name: Print Neatly. You will lose point if I cannot read your name or perm number. Perm Number: All work must be shown on the exam for partial credit. Points will be taken off for incorrect or no units. Calculators are allowed. Cell phones may not be used for calculators. On fundamental and short answer problems you must show your work in order to receive credit for the problem. If your cell phone goes off during the exam you will have your exam removed from you. Fundamentals (of 36 possible) ( )+( ) = Problem (of 6 possible) Problem 2 (of 8 possible) Multiple Choice (of 30 possible) Midterm Total (of 00 possible)
2 Fundamental Questions Each of these fundamental chemistry questions is worth 6 points. You must show work to get credit. Little to no partial credit will be awarded. Make sure to include the correct units on your answers. a) 4 pts A chemistry graduate student is studying the rate of this reaction: 2H3PO4(aq) P2O5(aq) + 3H2O(aq) He fills a reaction vessel with H3PO4 and measures its concentration as the reaction proceeds: Use this data to answer the following questions. Write the rate law for this reaction. (You must show your work to get credit) rate = k Time (minutes) [H3PO4] M M M M M Time (m) [H3PO4] ln[h3po4] The graph of t vs [H 3 PO 4 ] is a straight line therefore, rate=k[h3po4]2 b) 2 pts. Calculate the value of the rate constant k. Round your answers to 2 significant digits. Also be sure your answer has the correct unit symbol. = kt + therefore the slope equal k [A] [A] 0 slope = rise run = M M = 5.8 L 40. m 0 m m Or L s 2) 6 pts What mass of I2 from KI can be produced in.0 h with a current of 5 A? n = It (5 A) F = (.0 h( 96,486 C 2I - I 2 + 2e - 60 m s h )(60 m )) = 0.56 e 0.56 e ( I 2 2 e ) ( g I 2 I 2 ) = 7 g I 2 [H 3 PO 4 ] 3) 6 pts Balance N2H4(aq) + CO3 2- (aq) N2(g) + CO(g) in basic conditions N2H4 N2 + 4H + + 4e - CO H + + 2e - CO + 2H2O N2H4 N2 + 4H + + 4e - 2(CO H + + 2e - CO + 2H2O) N2H4 +2CO H + N2 + 2CO + 4H2O N2H4 +2CO H + + 4OH - N2 + 2CO + 4H2O + 4OH - N2H4(aq) + 2CO3 2- (aq) N2(g) + 2CO(g) + 4OH - (aq) 2
3 4) 6 pts Which of the reaction profiles represents the effect of a catalyst on the rate of reaction? A is the catalysized reaction. 5) 6 pts Draw the following galvanic cell: Pb(s) + Cl2(g) Pb 2+ (aq) + 2Cl - (aq) Calculate E, show the direction of electron flow and the direction of ion migration through the salt bridge, and identify the cathode and anode. Pb Pb e - E = 0.3 V Cl2 + 2e - 2Cl - E =.36 V Cl2(g) + Pb(s) Pb 2+ (aq) + 2Cl - (aq) E =.49 V 6) 6 pts In the first-order reaction A 2B, when the initial concentration of A was M it took 95 s for the concentration to drop to M. How much more time would be needed for the concentration to drop by an additional M. st Order ln[a] = kt + ln[a] o Determine k ln( M) = k(95 s) + ln( M) k = s Determine time ln( M) = (0.002 )t + ln( M) s t = 65s 3
4 Short Answer Questions Each of the following short answer questions are worth the noted points. Partial credit will be given. You must show your work to get credit. Make sure to include proper units on your answer. a) 6 pts Calculate the voltage of the following cell at 298 K Zn(s) Zn 2+ (aq).7 M Ag + (aq) 2.4 M Ag(s) Reaction of Interest Zn Zn e - E = 0.76 V 2(Ag + + e - Ag) E = 0.80 V Zn(s) + 2Ag + (aq) Zn 2+ (aq) + 2Ag(s) E =.56V E = E RT nf ln(q) = E RT nf ln ] ([Zn2+ [Ag + ] 2) E =.56V (8.345 K )(298K) (2)(96,485 C ) ln ( ) =.58 V b) 0 pts What is the voltage for the same cell at 800 K Zn(s) Zn 2+ (aq).7 M Ag + (aq) 2.4 M Ag(s) Hint: Assume ΔH and ΔS are temperature independent. = 42 S Zn K, S Ag = 43 K, S Ag + = 73, K ΔH f(ag+ ) = 05 k Find ΔG at 800 K G = H T S Find ΔH H rxn = H f (Zn 2+ ) + 2 H f (Ag) H f (Zn) 2 H f (Ag + ) H rxn = 53 k + 2(0 k ) 0 k Find ΔS S rxn = S Zn S Ag S Zn 2S Ag + 2(05 k S rxn = 2 + 2(43 ) K K 42 2 (73 ) = 24 K K G = H T S = 363,000 (800K) ( 24 G = 92,000 G = nfe E = G nf 92,000 = (2)(96,485 C ) = V E = 0.995V (8.345 K )(800K) (2)(96,485 C ) ln ( ) =.04 V ) = 363 k K ) K 4
5 2a) 8 pts The reaction 5Br - (aq) + BrO3 - (aq) + 6H + (aq) 3Br2(l) + 3H2O(l) Is expected to obey the mechanism k BrO3 - (aq)+h + (aq) HBrO3(aq) fast equilibrium k k 2 HBrO3(aq)+H + (aq) H2BrO3 + (aq) fast equilibrium k 2 Br - (aq)+h2bro3 + (aq) k 3 (Br-BrO2)(aq)+H2O(l) (Br-BrO2)(aq)+4H + (aq)+4br - (aq) products Write the rate law for this reaction. slow fast rate = k 3 [Br ][H 2 BrO 3 + ] H2BrO3 + is an intermediate therefore, needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate. k 2 [HBrO 3 ][H + ] = k 2 [H 2 BrO 3 + ] [H 2 BrO 3 + ] = k 2[HBrO 3 ][H + ] k 2 rate = k 3k 2 [Br ][HBrO 3 ][H + ] k 2 HBrO3 is an intermediate therefore, needs to be eliminated from the rate law. Use equilibrium to eliminate the intermediate. k [BrO 3 ][H + ] = k [HBrO 3 ] [HBrO 3 ] = k [BrO 3 ][H + ] k rate = k 3k 2 k [Br ][BrO 3 ][H + ] 2 k k 2 = k[br ][BrO 3 ][H + ] 2 2b) 0 pts Does this mechanism fit with the following experimental data? You must find the experimental rate law to get credit. [Br - ]o (M) [BrO3 - ]o (M) [H + ]o (M) Initial Rate( ) General Rate Law: rate = k[br ] x [BrO 3 ] y [H + ] z From experiments and 2 when the concentration of Br - is double the rate goes up by a factor of 2 therefore st order with respect to Br - From experiments and 3 when the concentration of H + is double the rate goes up by a factor of 4 therefore, 2 nd order with respect to H + From experiments and 4 2 L s = k(.0 M)(.0 M)y (.0 M) 2 36 L s = k(.0 M)(2.0 M)y (.0 M) = (0.50) y y=.6 rate = k[br ][BrO 3 ].6 [H + ] 2 No the mechanism is not consistent. Ls 5
6 Multiple Choice Questions On the ParScore form you need to fill in your answers, perm number, test version, and name. Failure to do any of these things will result in the loss of point. Your perm number is placed and bubbled in under the ID number. Do not skip boxes or put in a hyphen; unused boxes should be left blank. Bubble in your test version (B) under the test form. Note: Your ParScore forms will not be returned to you, therefore, for your records, you may want to mark your answers on this sheet. Each multiple choice question is worth 5 points. Identify the choice that best completes the statement or answers the question. Reaction E (volts) Na + + e Na 2.7 Al e Al.66 Fe e Fe 0.44 Co e Co 0.28 Cu e Cu Ag + + e Ag Cl2 + 2e 2Cl +.36 F2 + 2e 2F Which of the following would be the best reducing agent? a. Na + b. Al 3+ c. F2 d. Al e. F 2. The equilibrium constant for the following general reaction is 0. at 25 C. Calculate E for the cell. X2(s) + Y + (aq) X 2+ (aq) + Y(s) (unbalanced) a V b V c V d V 6
7 3. What time is required to plate a metal tray (24.0 cm x 2.0 cm) with a coating (thickness cm) of silver (density = 0.54 ) using a current of 7.65 A? Neglect the amount of silver required to coat the edges. Helpful Information: The reaction of interest is Ag + + e - Ag and MAg = 07.9 a. 30 s b. 708 s c. 92 s d. 420 s 4. The rate constant for a reaction increases from 0.0 to 00. when the temperature is increased from 35 K to 43 K. What is the activation energy for the reaction in? a b. 8.8 c..0 d In which of the following cases must E be equal to zero? I. In any cell at equilibrium II. In a concentration cell III. E can never be equal to zero. a. II only b. III c. I and II d. I only 6. What reaction occurs at the anode and at the cathode when electrolysis of aqueous MgI2 occurs? Assume standard conditions. Helpful information: I2 + 2e - 2I- E = 0.54 V Mg e - Mg E = V Anode/Cathode a. Water/Magnesium b. Water/Water c. Magnesium/Water d. Iodine/Magnesium 7
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